On Thu, Aug 23, 2007 at 06:22:52PM +0200, Mlynco Mlynco wrote: [..] > > So, ie I am looking for a recipe. I have indexed some of them with such > ingredients: > > recipe1: "beef" > recipe2: "onion beef chicken" > recipe3: "onion beef chicken tomato" > > Looking for a "beef" I wouldn't like to "punish" recipe2 and recipe3 > because they are richer, I would like to treat them in the same way. > Actually the score does not have to be 1.0, but I would like it to be > the same for all recipes mentioned above. > > Any suggestions?
have a look at ConstantScoreQuery. It allows to turn a Filter into a query where all results score equally. To get this Filter, you could use the QueryFilter class, which allows you to turn your original query into a Filter. query = ConstantScoreQuery.new(QueryFilter.new(TermQuery.new(:ingredients, 'beef')) looks like there should be an easier way to achieve this, though ;-) Jens -- Jens Krämer http://www.jkraemer.net/ - Blog http://www.omdb.org/ - The new free film database _______________________________________________ Ferret-talk mailing list [email protected] http://rubyforge.org/mailman/listinfo/ferret-talk
